Exercise
Given the differentiable function
z(x,y)=\ln\frac{1}{\sqrt{x^2+y^2}}
Prove the equation
z''_{xx}+z''_{yy}=0
Proof
We will use the chain rule to calculate the partial derivatives of z.
z'_x=\frac{1}{\frac{1}{\sqrt{x^2+y^2}}}\cdot\frac{-1}{{(\sqrt{x^2+y^2})}^2}\cdot\frac{1}{2\sqrt{x^2+y^2}}\cdot (2x)=
=\sqrt{x^2+y^2}\cdot\frac{-x}{x^2+y^2}\cdot\frac{1}{\sqrt{x^2+y^2}}=
=\frac{-x}{x^2+y^2}
z'_y=\frac{1}{\frac{1}{\sqrt{x^2+y^2}}}\cdot\frac{-1}{{(\sqrt{x^2+y^2})}^2}\cdot\frac{1}{2\sqrt{x^2+y^2}}\cdot (2y)=
=\sqrt{x^2+y^2}\cdot\frac{-y}{x^2+y^2}\cdot\frac{1}{\sqrt{x^2+y^2}}=
=\frac{-y}{x^2+y^2}
We will calculate the second order derivatives.
z''_{xx}=\frac{-(x^2+y^2)+x\cdot 2x}{{(x^2+y^2)}^2}=
=\frac{x^2-y^2}{{(x^2+y^2)}^2}
z''_{yy}=\frac{-(x^2+y^2)+y\cdot 2y}{{(x^2+y^2)}^2}=
=\frac{y^2-x^2}{{(x^2+y^2)}^2}
We will put the partial derivatives in the left side of the equation we need to prove.
z''_{xx}+z''_{yy}=
=\frac{x^2-y^2}{{(x^2+y^2)}^2}+\frac{y^2-x^2}{{(x^2+y^2)}^2}=
=\frac{x^2-y^2+y^2-x^2}{{(x^2+y^2)}^2}=
=\frac{0}{{(x^2+y^2)}^2}=
=0
We got zero as required.
Have a question? Found a mistake? – Write a comment below!
Was it helpful? You can buy me a cup of coffee here, which will make me very happy and will help me upload more solutions!